Solving for Displacement in Circular Motion

[shawonna23]

when using the sine law am I trying to find c?

 

[shawonna23]

Is it sin30/1.6=sin120/c

 

[christinono]

Here is the formula:
$$\frac{Sin A}{a} = \frac{Sin B}{b}$$

Now, place values in:

$$\frac{sin 30}{1.6km} = \frac{sin 120}{x}$$

isolate x:

$$x = \frac{sin 120}{sin 30} \times 1.6 km$$

Now, solve for x.

 

[shawonna23]

x=2.77 Now what do I do now?

 

[christinono]

What do you mean, that's the answer (I hope). Does you texbook or you worksheet give the answer?

 

[MathStudent]

thats the same answer i got ...

 

[shawonna23]

no it doesn't. So this is the answer for the magnitude. What is the direction in degrees?

 

[christinono]

Look back at the triangle.

 

[shawonna23]

Are you guys still there? What do I do to find the direction (relative to due east) in degrees?

 

[shawonna23]

Would I do this: theta= tan-1 (opp)/(adj)= 45 degrees?

 

[christinono]

Are U looking at you circle with the triangle in it? The angle U are looking for is the angle between the line that connects the 2 dots (initial and final positions of couple) and the line that connects the first dot to the origin.

 

[christinono]

Gotta run... :zzz: (actually, got to sleep)

BTW, the direction is 30 degrees North of East.

 

[shawonna23]

thanks so much

 

[christinono]

Glad I could help. :!)